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A spherical surface of radius of curvature \(R\) separates air (refractive index \(1.0\) ) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object \(P\) placed in air is found to have a real image \(Q\) in the glass. The line \(P Q\) cuts the surface at a point \(O\), and \(P O=O Q\). The distance \(P O\) is equal to (A) \(5 R\) (B) \(3 R\) (C) \(2 R\) (D) \(1.5 R\)

Short Answer

Expert verified
\(P O = R\)

Step by step solution

01

Recall and write down the lens maker's formula

The lens maker's formula is given by the equation: \(\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\) Where f is the focal length, n is the refractive index of the glass, and R1 and R2 are the radii of curvature of the two surfaces. Since the center of curvature is in the glass, R1 = R, and the second surface is flat (infinite radius), so R2 = ∞.
02

Simplify the lens maker's formula given the problem

Plugging the given values into the equation, we get: \(\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right)\) Simplifying this equation, we obtain: \(\frac{1}{f} = 0.5\left(\frac{1}{R}\right)\)
03

Use the thin lens formula and the given conditions

The thin lens formula is given by the equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Where do is the object distance from the surface and di is the image distance. From the problem, we know that PO = OQ, and PO = d_o. Hence, d_o = d_i.
04

Substitute the values in the thin lens formula

Substituting do = di in the thin lens formula, we get: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_o} \) \( \frac{1}{f} = \frac{2}{d_o} \)
05

Equate the thin lens formula and the lens maker's formula

Now, equate the two equations for \(1/f\): \(\frac{2}{d_o} = 0.5\left(\frac{1}{R}\right)\)
06

Solve for the distance PO

Solve the equation for d_o: \(d_o = 2 \cdot 0.5R\) \(d_o = R\) Since PO = d_o, we get: \(P O = R\) The answer is (D) 1.5R.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lens Maker's Formula
The Lens Maker's Formula is an essential equation in optics, helping us calculate the focal length of a lens based on its shape and the refractive index of the material it's made from. The formula is expressed as follows:\[\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]Here,
  • \(f\) is the focal length of the lens.
  • \(n\) is the refractive index of the lens material relative to the surrounding medium (e.g., air).
  • \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
For a spherical lens, one surface is convex (curves outward), while the other surface can be either convex or concave (curves inward). In the given exercise, we assume one surface is spherical with a known radius of curvature, while the other is flat, leading to \(R_2 = \infty\) (since a flat surface means the radius is infinitely large). With this simplification, the Lens Maker's Formula reduces to consider just \(R_1\) and has a more straightforward application.The formula forms the backbone of many optics problems, forming a bridge between theoretical physics and practical applications like eyeglasses, cameras, and microscopes. Understanding this equation helps not only in calculating focal lengths but also in designing lenses with desired refraction characteristics.
Refractive Index
The refractive index is a crucial concept in optics that describes how light bends when it enters a different medium. It is denoted by the symbol \(n\) and is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium:\[n = \frac{c}{v}\] Where
  • \(c\) is the speed of light in a vacuum (approximately \(3.00 \times 10^8\) m/s).
  • \(v\) is the speed of light in the given medium.
The refractive index indicates the degree to which light is refracted, or bent, as it transitions from one medium to another. A higher refractive index means that light travels more slowly in the medium and is bent more significantly upon entering or leaving the medium. In the exercise, air has a refractive index of \(1.0\) and glass has a refractive index of \(1.5\), indicating that light will slow down and bend toward the normal when entering glass from air. This is essential when determining focal points and image formation by lenses, as it influences how light rays converge or diverge. Understanding the refractive index is crucial for explaining phenomena like refraction, dispersion, and the critical angle.
Thin Lens Formula
The Thin Lens Formula is integral for understanding how lenses form images, combining both object and image distances with focal length. It is represented as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Where
  • \(f\) represents the focal length of the lens.
  • \(d_o\) is the distance from the object to the lens.
  • \(d_i\) is the distance from the image to the lens.
This equation establishes the relationship between them, allowing us to determine any unknown distance if the other two are given. It is particularly useful when dealing with "thin" lenses - lenses whose thickness is negligible compared to their focal length and curvature radii, thus simplifying the geometric analysis.In the exercise provided, the object and image distance are equal, leading to a simplified formula \(\frac{1}{f} = \frac{2}{d_o}\). Solving this establishes conditions for the given optical scenario, such as transforming image attributes like magnification or determining focal points. The Thin Lens Formula is a practical tool in predicting how lenses affect light paths, making it invaluable in fields ranging from photography to astronomy.

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Most popular questions from this chapter

Focal length of an equiconvex lens is \(20 \mathrm{~cm}\). If we cut it once perpendicular to principle axis, and then along principal axis, then focal length of each part will be (A) \(20 \mathrm{~cm}\) (B) \(10 \mathrm{~cm}\) (C) \(40 \mathrm{~cm}\) (D) \(5 \mathrm{~cm}\)

The magnifying power of an astronomical telescope in normal adjustment is 8 and the distance between the two lenses is \(54 \mathrm{~cm}\). The focal length of eye lens and objective lens will be, respectively. (A) \(6 \mathrm{~cm}\) and \(48 \mathrm{~cm}\) (B) \(48 \mathrm{~cm}\) and \(6 \mathrm{~cm}\) (C) \(8 \mathrm{~cm}\) and \(64 \mathrm{~cm}\) (D) \(64 \mathrm{~cm}\) and \(8 \mathrm{~cm}\)

When an object is placed at a distance of \(25 \mathrm{~cm}\) from a mirror, the magnification is \(m_{1} .\) The object is moved \(15 \mathrm{~cm}\) away with respect to the earlier position along principal axis, magnification becomes \(m_{2}\). If \(m_{1} \times m_{2}=4\), the focal length of the mirror is (A) \(10 \mathrm{~cm}\) (B) \(30 \mathrm{~cm}\) (C) \(15 \mathrm{~cm}\) (D) \(20 \mathrm{~cm}\)

If we put \(v=V+f\) and \(u=U+f\), the mirror formula \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) becomes (A) \((V+f)(U+f)=f^{2}\) (B) \(V U=f^{2}\) (C) \((V-f)(U-f)=f^{2}\) (D) \(V U=2 f^{2}\)

An object is placed in front of a convex mirror at a distance of \(50 \mathrm{~cm}\). A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is \(30 \mathrm{~cm}\), there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror (in \(\mathrm{cm}\) ) is (A) 60 (B) 50 (C) 30 (D) 25

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